2013 AMC 8 Problems/Problem 23
Contents
[hide]Problem
Angle of is a right angle. The sides of are the diameters of semicircles as shown. The area of the semicircle on equals , and the arc of the semicircle on has length . What is the radius of the semicircle on ?
Video Solution
https://youtu.be/crR3uNwKjk0 ~savannahsolver
Solution 1
If the semicircle on were a full circle, the area would be .
, therefore the diameter of the first circle is .
The arc of the largest semicircle is , so if it were a full circle, the circumference would be . So the .
By the Pythagorean theorem, the other side has length , so the radius is
~Edited by Theraccoon to correct typos.
Brief Explanation
SavannahSolver got a diameter of 17 because the given arc length of the semicircle was 8.5π. The arc length of a semicircle can be calculated using the formula πr, where r is the radius. let’s use the full circumference formula for a circle, which is 2πr. Since the semicircle is half of a circle, its arc length is πr, which was given as 8.5π. Solving for r, we get 𝑟=8.5 . Therefore, the diameter, which is 2r, is 2x8.5=17 Then, the other steps to solve the problem will be the same as mentioned above by SavannahSolver the answer is
. - TheNerdWhoIsNerdy.
Solution 2
We go as in Solution 1, finding the diameter of the circle on and . Then, an extended version of the theorem says that the sum of the semicircles on the left is equal to the biggest one, so the area of the largest is , and the middle one is , so the radius is .
~Note by Theraccoon: The person who posted this did not include their name.
Video Solution by OmegaLearn
https://youtu.be/abSgjn4Qs34?t=2584
~ pi_is_3.14
See Also
2013 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 22 |
Followed by Problem 24 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AJHSME/AMC 8 Problems and Solutions |
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